Tìm số nguyên x, y biết:
a) (x+ 5).(3x- 12) >0
b) (x3+ 5).(x3 +10).(x3+ 15).(x3+ 30) <0
c) xy= x+ y
d) xy+ 12= x+ y
Tìm x biết:
a) x3 - 7x2 - 9x + 63 = 0
b) x3 - 3x2 + 3x - 1 + 2.(x2 - x) = 0
Bài 5: Giải các phương trình sau:
a. (3x - 1)2 - (x + 3)2 = 0
b. x3 = \(\dfrac{x}{49}\)
c. x2 - 7x + 12 = 0
d. 4x2 - 3x -1 = 0
e. x3 - 2x - 4 = 0
f. x3 + 8x2 + 17x +10 = 0
g. x3 + 3x2 + 6x + 4 = 0
h. x3 - 11x2 + 30x = 0
a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
*Cách khác:
a) Ta có: \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(3x-1\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-x-3\\3x-1=x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-2\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{2};2\right\}\)
Bài 1: Số(−3)20+1(−3)20+1 có phải là tích của hai số nguyên liên tiếp không?
Bài 2: Tìm x∈Zx∈Z biết (x+5)x (3x-12)>0
Bài 3: Tìmx∈Zx∈Z biết (x3+5)(x3+10)(x3+15)(x3+30)<0
Bài 3 (2đ): Tìm x biết:
a. (x - 8 )( x3+ 8) = 0
b. (4x - 3) – ( x + 5) = 3(10 - x)
\(a.\)
\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
\(S=\left\{8,-2\right\}\)
\(b.\)
\(\left(4x-3\right)-\left(x+5\right)=3\cdot\left(10-x\right)\)
\(\Leftrightarrow4x-3-x-5-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{38}{6}\)
\(S=\left\{\dfrac{38}{6}\right\}\)
a) \(\left(x-8\right)\left(x^3+8\right)=0\)
=>\(x-8=0 => x=8\)
hoặc \(x^3+8=0\)=>\(x=-2\)
b) \(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\)
\(< =>3x-8=3\left(10-x\right)\)
\(< =>3x-8-30+3x=0\)
\(< =>6x=38=>x=\dfrac{38}{6}=\dfrac{19}{3}\)
Bài 3 (2đ): Tìm x biết:
a) (x - 8 )( x3 + 8) = 0
b) (4x - 3) – ( x + 5) = 3(10 - x)
a) (x - 8 )( x3 + 8) = 0
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x^3=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
b)(4x - 3) – ( x + 5) = 3(10 - x)
\(\Leftrightarrow4x-3-x-5=30-3x\)
\(\Leftrightarrow3x-8=30-3x\)
\(\Leftrightarrow3x-8-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{19}{3}\)
Sửa lại câu `b) :`
`a)`
`( x-8 )( x^3 + 8 )`
`=> x-8=0` hoặc `x^3+8=0`
`=> x=8` hoặc `x^3 = -8=(-2)^3`
`=> x=8` hoặc `x=-2`
Vậy `x in { -2;8}`
`b)`
`( 4x-3 ) - ( x+5) = 3( 10-x)`
`=> 4x-3-x-5=30-3x`
`=> ( 4x-x)+(-3-5)=30-3x`
`=> 3x-8=30-3x`
`=> 6x=38`
`=> x=19/3`
Vậy `x=19/3`
`a)`
`( x-8 )( x^3 + 8 )`
`=> x-8=0` hoặc `x^3+8=0`
`=> x=8` hoặc `x^3 = -8=(-2)^3`
`=> x=8` hoặc `x=-2`
Vậy `x in { -2;8}`
`b)`
`( 4x-5 ) - ( x+5) = 3( 10-x)`
`=> 4x-5-x-5=30-3x`
`=> ( 4x-x)+(-5-5)=30-3x`
`=> 3x-10=30-3x`
`=> 6x=40`
`=> x=20/3`
Vậy `x=20/3`
Tìm x ϵ Z để ( x3+5)( x3+10)(x3+15)(x3+30) <0
\(TH_1:x\ge0\Leftrightarrow x^3\ge0\Leftrightarrow VT>0\left(loại\right)\)
\(TH_2:x< 0\)
Với \(x=-1\Leftrightarrow VT=4\cdot9\cdot14\cdot29>0\left(loại\right)\)
Với \(x=-2\Leftrightarrow VT=-3\cdot2\cdot7\cdot23< 0\left(nhận\right)\)
Với \(x=-3\Leftrightarrow VT=-22\left(-17\right)\left(-12\right)\cdot3< 0\left(nhận\right)\)
Với \(x< -4\Leftrightarrow x^3< -64\Leftrightarrow x^3+5< x^3+10< x^3+15< x^3+30< 0\)
Do đó cả 4 thừa số trong tích đều âm nên tích này luôn dương
Vậy \(x\in\left\{-2;-3\right\}\)
Bài 5. Tìm x, biết:
a) x (2x - 7) + 4x -14 = 0
b) x3 - 9x = 0
c) 4x2 -1 - 2(2x -1)2 = 0
d) (x3 - x2 ) - 4x2 + 8x - 4 = 0
\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Bài 2 (2 đ): Cho các đa thức sau:
P(x) = x3 – 6x + 2
Q(x) = 2x2 - 4x3 + x - 5
a) Tính P(x) + Q(x)
b) Tính P(x) - Q(x)
Bài 3 (2đ): Tìm x biết:
a. (x - 8 )( x3+ 8) = 0
b. (4x - 3) – ( x + 5) = 3(10 - x)
Bài 2
P(x) + Q(x) = x3 – 6x + 2 + 2x2 - 4x3 + x - 5 = - 3x3 + 2x2 – 5x - 3
P(x) - Q(x) = x3 – 6x + 2 - 2x2 + 4x3 - x + 5 = 5x3 − 2x2 − 7x+7
Bai 3
a)(x-8)(x3+8)=0
=>x-8=0 hoac x3+8=0
=>x =8 hoac x3 =-8
=>x =8 hoac x =-2
Vậy x=8 hoặc x=-2
b)(4x-3)-(x+5)=3(10-x)
=>4x-3-x-5=30-3x
=>4x-x+3x=30+3+5
=>x(4-1+3)=38
=>6x =38
=>x =\(\dfrac{38}{6}\)
=>x =\(\dfrac{19}{3}\)
Vậy x=\(\dfrac{19}{3}\)
Bài 1: Tìm x, biết:
a) 3x2-3x+2x3-2x2=0
b) x3+27=-x2+9
a)\(3x\left(x-1\right)+2x^2\left(x-1\right)=0\\ \Leftrightarrow x\left(x-1\right)\left(3+2x\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\3+2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=\dfrac{-3}{2}\end{matrix}\right.\)
a: Ta có: \(3x^2-3x+2x^3-2x^2=0\)
\(\Leftrightarrow2x^3+x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+x-3\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b: Ta có: \(x^3+27=-x^2+9\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-3\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
Tìm x biết:
a) 27x3-54x2+36x-8=0
b) x3+15x2+75x+125=0
c) x3-18x2+108x-216=0
d) (x-1)3-x.(x2+3x)=2
e) (x+1)3+(x-2)3-2x2.(x-1,5)=3
Giải chi tiết giúp mình nha.Cảm ơn nhiều